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Today I've got an insteresting question about geometry. Let's get into it.

Let $ABC$ be a triangle such that $AC$ is its shortest side. A point $P$ is inside it such that $BP = AC.$ Let $R$ be the midpoint of $BC$ and let $M$ be the midpoint of $AP.$ $E$ is the intersection of $BP$ with $AC.$ Show that the bisector of $\angle BEA$ is perpendicular to line $MR$

Is there any idea to go along this type of perpendicularity problems?

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    $\begingroup$ What have you tried? $\endgroup$
    – Math Lover
    Commented Dec 29, 2021 at 20:39
  • $\begingroup$ I've tried to use analytic geometry and also geometry but I couldn't get further. I really don't have an specfic tactic with that problem, instead I want to share and get ideas with someone else $\endgroup$
    – oscar AMVS
    Commented Dec 29, 2021 at 21:45

2 Answers 2

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Triangle ABC

Let the midpoint of $AB$ be $D$. Now, $RD=\frac {1}{2} PB=\frac {1}{2} CA=DM$ and $\angle MDR=180^\circ-\angle ADR-\angle MDB=180^\circ-\angle ABE-\angle BAE=\angle BEA.$

The angle bisectors of $\angle RDM$ and $\angle EAB$ are parallel since they are equal and $MD\parallel AC$. Since the angle bisector of $\angle RDM$ is perpendicular on $MR$ ($\triangle RDM$ is isosceles), the angle bisector of $\angle BEA$ is perpendicular on $MR$.

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You can find the equation of the line $BF$ where $F$ is the midpoint of the segment $AE$ (that is the bisector of the angle $\angle BEA$).

Then you can "find" $M$ by assigning a value to $P$ (e.g: $(x, y)$).

The you find the equation for the line $MR$.

The lines are perpendicular if $f'(x) = -g'(x)$ where $f$ and $g$ are the equations for the lines.

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